Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $n = \dfrac{4q + 8}{q + 1} \div \dfrac{q^3 - 16q^2 + 63q}{q^3 - 8q^2 - 9q} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{4q + 8}{q + 1} \times \dfrac{q^3 - 8q^2 - 9q}{q^3 - 16q^2 + 63q} $ First factor out any common factors. $n = \dfrac{4(q + 2)}{q + 1} \times \dfrac{q(q^2 - 8q - 9)}{q(q^2 - 16q + 63)} $ Then factor the quadratic expressions. $n = \dfrac {4(q + 2)} {q + 1} \times \dfrac {q(q - 9)(q + 1)} {q(q - 9)(q - 7)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac {4(q + 2) \times q(q - 9)(q + 1) } {(q + 1) \times q(q - 9)(q - 7) } $ $n = \dfrac {4q(q - 9)(q + 1)(q + 2)} {q(q - 9)(q - 7)(q + 1)} $ Notice that $(q - 9)$ and $(q + 1)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {4q\cancel{(q - 9)}(q + 1)(q + 2)} {q\cancel{(q - 9)}(q - 7)(q + 1)} $ We are dividing by $q - 9$ , so $q - 9 \neq 0$ Therefore, $q \neq 9$ $n = \dfrac {4q\cancel{(q - 9)}\cancel{(q + 1)}(q + 2)} {q\cancel{(q - 9)}(q - 7)\cancel{(q + 1)}} $ We are dividing by $q + 1$ , so $q + 1 \neq 0$ Therefore, $q \neq -1$ $n = \dfrac {4q(q + 2)} {q(q - 7)} $ $ n = \dfrac{4(q + 2)}{q - 7}; q \neq 9; q \neq -1 $